Multiple of a number and sum of a set Lets say there is a number 30, and this sum of numbers 
3 + 5 + 7 + 9 ... 

Now I want to find where this sum and a multiple of 30 gets equal.
for 30 its 4 = 30 * 4 = 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
I don't have any math experience, just wanted to know if this could be calculated by some formula.
I can imagine, that it must somehow be related to circles/sine-waves, where 30 is the length of the circle and the sum of this numbers must somehow match a multiple of it.
Maybe there is some kind of formula for this already, would like to know, thanks
PS: would be nice, if you could also put the right tags to this question
 A: We know that the sum of first $n$ odd numbers is $n^2$ . So the sum $$1+3+5+\cdots+(n+1) = n^2 \\ \implies 3+5+7+\cdots+(n+1)=(n+1)^2-1$$Since we want it to be a multiple of $30$ , We have the following equation:
$$(n+1)^2- 1= 30k \implies n^2+2n-30k = 0$$
Using quadratic formula , we get:
$$n = \dfrac{-2\pm\sqrt{4+120k}}2\implies n = -1+\sqrt{1+30k}$$
Since $n$ is an integer , it follows that $(1+30k)$ must be a perfect square of an odd number.
$$1+30k = (2x+1)^2 \implies k = \dfrac{2x(x+1)}{15}$$
Since $2$ is not divisible by $15$ , it follows that $x^2+x\equiv 0\mod 15$
and we get the possible values of $x \equiv 0,5,9,14 \mod 15$ $($The proof is left to the readers .$)$ 
Our previous equation becomes , 
$$n = -1 + 2x+1 \implies n = 2x$$
So for $x \equiv 0,5,9,14 \mod 15$ , you can get a possible solution $$(n,k) = \left(2x , \dfrac{2x(x+1)}{15}\right)$$
A: First you should know the sum of arithmetic series.  The
arithmetic series general term is $a_n=a_1 +(n-1)r,$ where $r$ is the base the constant you add with every term. 
The sum of the first $n$ terms is given by 
$$S_n= na_1+r(1+2+\cdots +(n-1))=na_1 +\frac{r n(n-1)}{2}=\frac{n(2a_1+r(n-1))}{2}.$$
In our case $a_1=3, r=2.$
So we want $n$ such that 
$$\frac{n(2a_1+r(n-1))}{2}=30k.$$ 
Hence 
$$\frac{n(4+2n)}{2}=30k$$
$$n(4+2n)=60k$$
$$n(2+n)=30k$$ we want first $n$ such that the left hand side is multiple of $30$ we know that $30$ contains $2,3,5$ so you need to make the left hand side contains these primes and $10$ is good because it contains $2,5$ and the $10+2$ contains the $3$ . If $n$ is 10 then $k=4$
Arthimatic series is a series begin with a number and you add a constant with each term ( the difference between conseqetive numbers is a constant ) . To find the formula for the sum we depend on the sum of $1+2+3+\cdots +n=\frac{n(n-1)}{2}$ 
